Comments on: Invertibility of random matrices http://www.tangentspace.net/cz/archives/2007/12/invertibility-of-random-matrices/ somewhere near the beginning. Fri, 21 Nov 2008 18:39:51 +0000 http://wordpress.org/?v=2.5 By: Alex http://www.tangentspace.net/cz/archives/2007/12/invertibility-of-random-matrices/#comment-239234 Alex Mon, 24 Dec 2007 06:27:13 +0000 http://www.tangentspace.net/cz/archives/2007/12/invertibility-of-random-matrices/#comment-239234 For your first point, there isn't any cheating: the rand call generates a matrix by choosing random i.i.d. [tex]U(0,1)[/tex] entries, then the orth call runs some orthogonalizing procedure on them -- you only get back a square matrix if the original matrix was full rank. I think this is equivalent to the procedure you describe. Also, selecting a random point then projecting it onto the sphere seems like a good way to select a random point in [tex]S^{n-1}[/tex]. For the second, yep. If you have [tex]k < n-1[/tex] vectors, then the probability of selecting a vector that lies in the span of those is the measure of the (appropriate section of the) hyperplane they define over the measure of the hypercube: 0. For your first point, there isn’t any cheating: the rand call generates a matrix by choosing random i.i.d. U(0,1) entries, then the orth call runs some orthogonalizing procedure on them — you only get back a square matrix if the original matrix was full rank. I think this is equivalent to the procedure you describe. Also, selecting a random point then projecting it onto the sphere seems like a good way to select a random point in S^{n-1}.

For the second, yep. If you have k < n-1 vectors, then the probability of selecting a vector that lies in the span of those is the measure of the (appropriate section of the) hyperplane they define over the measure of the hypercube: 0.

]]> By: ObsessiveMathsFreak http://www.tangentspace.net/cz/archives/2007/12/invertibility-of-random-matrices/#comment-239206 ObsessiveMathsFreak Mon, 24 Dec 2007 03:08:28 +0000 http://www.tangentspace.net/cz/archives/2007/12/invertibility-of-random-matrices/#comment-239206 Sorry for the double post. I forgot. In answer to your second question, the odds of getting a full ranked matrix are pretty high. It's virtually impossible for a random matrix in R^(nxn) to be singular.... I think. Sorry for the double post. I forgot. In answer to your second question, the odds of getting a full ranked matrix are pretty high. It’s virtually impossible for a random matrix in R^(nxn) to be singular…. I think.

]]> By: ObsessiveMathsFreak http://www.tangentspace.net/cz/archives/2007/12/invertibility-of-random-matrices/#comment-239205 ObsessiveMathsFreak Mon, 24 Dec 2007 03:06:50 +0000 http://www.tangentspace.net/cz/archives/2007/12/invertibility-of-random-matrices/#comment-239205 The matlab code could be cheating by getting a random matrix, checking for full rank, then orthonormalising it. The right way to do this in my opinion would be to pick a random point on S^(n-1), then pick another and orthonormalise, then another, etc, etc, until you have a full set of orthonormal vectors. However, I'm not sure how one would actually go about properly selecting a random point on S^(n-1). The matlab code could be cheating by getting a random matrix, checking for full rank, then orthonormalising it. The right way to do this in my opinion would be to pick a random point on S^(n-1), then pick another and orthonormalise, then another, etc, etc, until you have a full set of orthonormal vectors. However, I’m not sure how one would actually go about properly selecting a random point on S^(n-1).

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